A three-span continuous beam has an internal hinge at B. Section B is at the mid-span of AC. Section E is at the mid-span of CG. The 20 kN load is applied at section B whereas 10 kN loads are applied at sections D and F as shown in the figure. Span GH is subjected to uniformly distributed load of magnitude 5 kN/m. For the loading shown, shear force immediate to the right of section E is 9.84 kN upwards and the hogging moment at section E is 10.31 kN-m.

AB = BC = 2 m ; CD = DE = EF = FG = 1 m; GH = 4 m

The magnitude of the shear force immediate to the left and immediate to the right of section B are, respectively

0 & 20 kN

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10 kN & 10 kN

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20 kN & 0

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9.84 kN & 10.16 kN

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Solution

The correct option is A 0 & 20 kN

AB = BC = 2m

CD = DE = EF = FG = 1m

GH = 4m

B is internal hinge So

$\sum M = 0$

$R_{A} \times A B = 0$

$\Rightarrow$ $R_{A} = 0$

Shear force immediate to the left of B

$R_{A} = 0$

Shear force immediate to the right of B

$R_{A} - 20 = 20 k N$

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Q. A three-span continuous beam has an internal hinge at B. Section B is at the mid-span of AC. Section E is at the mid-span of CG. The 20 kN load is applied at section B whereas 10 kN loads are applied at sections D and F as shown in the figure. Span GH is subjected to uniformly distributed load of magnitude 5 kN/m. For the loading shown, shear force immediate to the right of section E is 9.84 kN upwards and the hogging moment at section E is 10.31 kN-m.

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The correct option is A 0 & 20 kN AB = BC = 2m CD = DE = EF = FG = 1m GH = 4m B is internal hinge So ∑M=0 RA×AB=0 ⇒ RA=0 Shear force immediate to the left of B RA=0 Shear force immediate to the right of B RA−20=20kN

The correct option is A 0 & 20 kN

AB = BC = 2m

CD = DE = EF = FG = 1m

GH = 4m

B is internal hinge So

$\sum M = 0$

$R_{A} \times A B = 0$

$\Rightarrow$ $R_{A} = 0$

Shear force immediate to the left of B

$R_{A} = 0$

Shear force immediate to the right of B

$R_{A} - 20 = 20 k N$